proof of general means inequality

Let w1, w2, …, wn be positive real numbers such that
w1+w2+⋯+wn=1. For any real numberr≠0, the
weighted power mean of degree r of n positive real numbers x1,
x2, …, xn (with respect to the weights w1, …,
wn) is defined as

Mwr⁢(x1,x2,…,xn)=(w1⁢x1r+w2⁢x2r+⋯+wn⁢xnr)1/r.

The definition is extended to the case r=0 by taking the limit
r→0; this yields the weighted geometric mean

Mw0⁢(x1,x2,…,xn)=x1w1⁢x2w2⁢…⁢xnwn

(see derivation of zeroth weighted power mean). We will prove the
weighted power means inequality, which states that for any two real
numbers r<s, the weighted power means of orders r and s
of n positive real numbers x1, x2, …, xn satisfy the
inequality

Mwr⁢(x1,x2,…,xn)≤Mws⁢(x1,x2,…,xn)

with equality if and only if all the xi are equal.

First, let us suppose that r and s are nonzero. We
distinguish three cases for the signs of r and s: r<s<0,
r<0<s, and 0<r<s. Let us consider the last case, i.e. assume
r and s are both positive; the others are similar. We write
t=sr and yi=xir for 1≤i≤n; this implies
yit=xis. Consider the function

with equality if and only if y1=y2=⋯=yn. By substituting
t=sr and yi=xir back into this inequality, we get

(w1⁢x1r+w2⁢x2r+⋯+wn⁢xnr)s/r≤w1⁢x1s+w2⁢x2s+⋯+wn⁢xns

with equality if and only if x1=x2=⋯=xn. Since s is
positive, the function x↦x1/s is strictly increasing, so
raising both sides to the power 1/spreserves the inequality:

(w1⁢x1r+w2⁢x2r+⋯+wn⁢xnr)1/r≤(w1⁢x1s+w2⁢x2s+⋯+wn⁢xns)1/s,

which is the inequality we had to prove. Equality holds if and only
if all the xi are equal.

If r=0, the inequality is still correct: Mw0 is defined as
limr→0⁡Mwr, and since Mwr≤Mws for all r<s with
r≠0, the same holds for the limit r→0. The same argument
shows that the inequality also holds for s=0, i.e. that
Mwr≤Mw0 for all r<0. We conclude that for all real numbers
r and s such that r<s,